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Suppose $a$ and $b$ are elements of a finite field of order $2^n$ with $n$ odd and $a^2+ab+b^2=0$. Is it necessary that both $a$ and $b$ must be zero ?

I understand that the field has characteristic $2$ but don't know how to use the fact that $n$ is odd, please help.

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If $b$ were not 0, say, then the ratio $a/b$ is a root of $x^2 + x + 1$ (divide the equation by $b^2$). What does that tell you about the subfield ${\mathbf F}_2(a/b)$ inside your finite field? – KCd Dec 22 '12 at 14:41
Well, if $n$ is $2$ for example you can take $b=1$ and $a$ to be either of the elements that are neither $1$ nor $0$. – Chris Eagle Dec 22 '12 at 14:41
@Chris: It was specified that $n$ is odd. – Cameron Buie Dec 22 '12 at 14:46
@CameronBuie: Yes, I know. The OP seemed to be unsure why this was relevant. – Chris Eagle Dec 22 '12 at 14:47
KCd nailed it. Another equivalent way to think about this is to determine the fields $GF(2^n)$ that contain a primitive third root of unity. – Jyrki Lahtonen Dec 22 '12 at 18:34
up vote 5 down vote accepted

If $b$ were not 0 then $a/b$ would be a root of $x^2 + x + 1$, which is irreducible over ${\mathbf F}_2$. Look at the size of the field ${\mathbf F}_2(a/b)$ and the size of the field you are working in that has order $2^n$.

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If $a=b$, then $a^2+ab+b^2=3a^2=a^2=0$ and $a=0=b$, we are done.

Suppose that $a\neq b$.

Observe that $0=(a-b)(a^2+ab+b^2)=a^3-b^3$. Thus, $a^3=b^3$.

We claim that $a=0$ and $b=0$.

If $a\neq 0$, then $(a^{-1}b)^3=1$ and the multiplicative order of $a^{-1}b$ in the multiplicative group $F-\{0\}$ is $1$ because $3\not\mid |F-\{0\}|=2^n-1$. Hence, $a^{-1}b=1$ and $a=b$, a contradiction. Therefore, $a=0$. By a similar argument, we have $b=0$.

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And $3\nmid 2^n-1$ because $n$ is odd. Right. – anon May 24 '15 at 2:21

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